# 8085 program to print the table of input integer

Problem – Write an assembly language program in 8085 to print the table of input integer.

Assumption – Suppose the inputted number is at memory location 2050 and the table will be printed from starting location 3050.

Example – Algorithm –

1. Load the value of input in accumulator from memory location 2050 and then copy it to another register say D.Also store 0A in register B.
2. Store memory location 3050 in M using LXI instruction and take another register say C with its value 00.
3. Now copy the content of D register to A and add the contents of A and C and store it in A then copy it to M.
4. Increment value of M by 1.
5. Copy content of A to C and decrements the content of B by 1 and if its value is 0 then halt otherwise again go to step number 3.

Program –

ADDRESS MNEMONICS COMMENTS
2000 LDA 2050 A<-
2003 MOV D, A D<-[A]
2004 MVI B 0A B<-0A
2006 LXI H 3050 H<-30 & L<-50
2009 MVI C 00 C<-00
200B MOV A, D A<-[D]
200C ADD C A<-[A]+[C]
200D MOV M, A M<-[A]
200E INX H HL<-HL+1
200F MOV C, A C<-[A]
2010 DCR B B<-[B]-1
2011 JNZ 200B Jump to address 200B if ZF=0
2014 HLT Terminates the program

Explanation –

1. LDA 2050: load the contents from 2050 memory location to accumulator(register A).
2. MOV D, A: move the contents of accumulator to register D.
3. MVI B 0A: store 0A data into register B.
4. LXI H 3050: store 30 in H register and 50 in L register, hence M will contain 3050 inside it.
5. MVI C 00: store 00 data in register C.
6. MOV A, D: move the contents of D register into A.
7. ADD C: add the contents of A and C register and store in A.
8. MOV M, A: move the contents of A register into M.
9. INX H: increments content of M by 1.
10. MOV C, A: move the contents of A register into C.
11. DCR B: decrements the content of B register by 1.
12. JNZ 200B: jump to address 200B if Carry flag is not zero.
13. HLT: terminate the program.

This article is attributed to GeeksforGeeks.org

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